exponents

I think answer should be A, if I understood Q correctly
(2^k)*(5^(k − 1))?

2*(2^(k-1))*(5^(k − 1))

2*10^(k-1)

OA?

OA - A. Thanks..

Is there another way to do this problem?

beater wrote:OA - A. Thanks..

Is there another way to do this problem?

but the q says (5^k − 1) and not (5^(k − 1))
is this a typo?

A. 2(10^k − 1)

Choice A must really be 2 * (10 ^(k-1))

= 2 * ((2*5) ^ (k-1))
= 2 * 2 ^ k-1 * 5 ^ k-1
= 2 ^ k * 5 ^ k-1

stop@800 wrote:

beater wrote:OA - A. Thanks..

Is there another way to do this problem?

but the q says (5^k − 1) and not (5^(k − 1))
is this a typo?

I suppose, yes
I was not able to get any offered answer from (2^k)(5^k − 1)
It should be
(2^k)*(5^(k − 1))?

i don't get it. can someone clarify pls?

so okay, back solving from answer A helps. but is there any straight cut way to do it?

i don't get it. can someone clarify pls?

as solved earlier ! key point here is to bring

2^k also in the form k-1 so that common factor can be take ouy

so below it has been done just that

2*(2^(k-1))*(5^(k − 1))

try solvoing it u wil get the question

2(2^k/2)*(5^(k − 1))


so going back to equation

2*(2^(k-1))*(5^(k − 1))

k-1 factor out.... powers being same base is multiplied
u get
2*10^k-1 which is one the option



so u get it now???

HTH
vishu[/quote]